Let $G$ be a graph of genus $g$. Is the number of (non necessarily disjoint) 5-clique subgraphs at most $f(g)$ for some function $f$?

Question

For a graph of genus $g$, it holds that it cannot have too many disjoint 5-cliques, as each clique requires a new handle. It feels that given a graph of genus $g$, it cannot have an unbounded number of non-disjoint 5-cliques either.

It's not really clear how to proceed. One might use the generalized Euler Formula to get $m \leq 3n-6+3g \implies \mathop{\mathrm{AverageDegree}}(G)=\frac{2m}{n} \leq 6+\frac{3g+6}{n}$, so for large graphs the average degree gets close to 6.0001, but this may not get us anywhere. It also holds that if $G$ can have an unbounded number of 5-cliques, an unbounded number of 5-cliques must share at least 1 vertex, else there will be an unbounded number of disjoint 5-cliques.

Answer 1

Yes, this is true. See my paper Subgraph densities in a surface with Gwenaël Joret and David Wood. We prove that for every $s \geq 5$, the number of $s$-cliques in a graph of Euler genus $g$ is at most $ \left( \frac{300 \sqrt{g}}{s} \right)^s$. See Theorem 6.9, where we give an almost matching lowerbound. We also determine the exact answer for small $s$ and $g$. See Table 1 in Section 6.4.

If you do not care about the precise bound, here is a reasonably self-contained proof.

Towards a contradiction, suppose that there is a graph $G$ embedded on a surface of Euler genus $g$ with many $5$-cliques. By the Sunflower lemma, $G$ contains a large family of $5$-cliques with the same pairwise intersection $K$. If $|K| \leq 2$, then we contradict the additivity of Euler genus. If $|K| \geq 3$, then $G$ contains a $K_{3,t}$ subgraph, with $t$ large. However, it is well-known that $K_{3,3g+2}$ does not embed on a surface of Euler genus $g$.

Answer 2

Say $\gamma(G)$ is the genus of a graph $G$. If $G$ has components $G_1,\dots,G_c$ then $\gamma(G)=\sum_{i=1}^c \gamma(G_i)$. This property is called the additivity of genus (and much stronger results are known). So if $G$ has $c$ disjoint copies of $K_5$, then $\gamma(G)\geq c\, \gamma(K_5)=c$ (by deleting vertices or edges not in any of the copies, and applying the above result). That is, the number of disjoint copies of $K_5$ is at most $\gamma(G)$.